# A problem with my reasoning in a problem about combinations

I was given the following problem to solve:

A committee of five students is to be chosen from six boys and five girls. Find the number of ways in which the committee can be chosen, if it includes at least one boy.

My method was $\binom{6}{1}\binom{10}{4}= 1260$, using the logic of choosing $1$ boy, then choosing the rest. This was wrong, as the answer was $\binom{11}{5}-\binom{5}{5}= 461$. The correct answer's logic was committee with no restrictions – committee with no girls.

• Velcome to our site! – kjetil b halvorsen Jun 17 '14 at 11:00

Your method has two errors. First you are selecting one boy from 6 and the other four from the whole set of 10 (girls and boys). Further you are only selecting one boy when you should select all combination of 1, 2, 3, 4, 5 boys. Instead circumventing this route, you select five people from 11 people and subtract the term with zero boys Thus

it is either: ${11\choose5} - {5\choose5}$ or

it is $${6\choose1}{5\choose4}+{6\choose2}{5\choose3}+{6\choose3}{5\choose2}+{6\choose4}{5\choose1}+{6\choose5}{5\choose0}$$

Thanks

Satish

• This is the right answer, but it doesn't explain why the OP's method was wrong. – David Jun 17 '14 at 10:52
• Thank you! Have another more complicated one going to ask soon :( – user9856 Jun 17 '14 at 10:54
• I had indicated the two errors that he made in his method. He seems to think you can select one boy from 6 and the other four from the rest of 10 (11-1). That is the first error. Second error is that he seems to have closed out the solution with one boy combination when he should have considered all combination of 1 boy, 2 boys,... 5 boys. – Satish Ramanathan Jun 17 '14 at 10:54

Your idea is to choose one boy and then four others, which might include further boys. Nice idea, but unfortunately it doesn't work: the reason why should be clear from the following choices.

1. Choose the boy $B_1$, then four more people $G_1,G_2,G_3,B_2$.
2. Choose the boy $B_2$, then four more people $G_1,G_2,G_3,B_1$.

These are two of the committees you have counted. . .

. . . BUT they are actually the same committee, so you should not have counted it twice.

Similarly, by following your method, a committee $B_1,B_2,B_3,G_1,G_2$ would be counted three times, and so on. This is why your method gives the wrong answer.

• Thank you! I see now why my idea was wrong... Is there a simple rule to remember concerning this type of questions? I keep finding myself using the wrong logic and ending up with the wrong answer... – user9856 Jun 17 '14 at 11:07
• Unfortunately I don't think there is a simple rule. It's largely a matter of experience, which is one of the reasons why counting problems can be difficult. But I would strongly recommend that whenever you solve a counting problem, you think very carefully about whether there is any possibility that you have counted the same outcome more than once. – David Jun 17 '14 at 11:11
• Okay, thanks again! :) – user9856 Jun 17 '14 at 11:14